hdu1495 bfs解决倒水问题

/*
1.每次倒水时的状态
cup: v1,v2,v3,step;
2.倒水共有6种情况，可通过双重循环实现
3.实现倒水函数
4.令v3恒小于v2,临界条件v1==v2且v3==0
*/
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAXN 105

using namespace std;

int vis[MAXN][MAXN][MAXN];//每一维分别表示：桶1,2,3中的水的体积。
int S, M, N;
int V[4];

//记录每次倒水后的状态
typedef struct node {
	int v[4];
	int step;
}note;

note tmp;

void pour(int i, int j)//水从第i桶向第j桶倒
{
	int sum = tmp.v[i] + tmp.v[j];//两桶水的总体积
	if (sum > V[j])//倒部分水
	{
		tmp.v[i] = tmp.v[i] - (V[j] - tmp.v[j]);
		tmp.v[j] = V[j];
	}
	else//全倒完
	{
		tmp.v[j] += tmp.v[i];
		tmp.v[i] = 0;
	}
}

void bfs()
{
	queue<note>	q;
	while (!q.empty())	q.pop();
	
	note t;
	t.step = 0; t.v[1] = S; t.v[2] = 0; t.v[3] = 0;
	q.push(t);
	vis[t.v[1]][t.v[2]][t.v[3]] = 1;

	while (!q.empty())
	{
		note cur = q.front(); q.pop();
		//临界条件
		if (cur.v[1] == cur.v[2] && cur.v[3] == 0)
		{
			printf("%d\n", cur.step);
			return;
		}
		//遍历6种倒水情况
		for (int i = 1; i <= 3; i++)
		{
			for (int j = 1; j <= 3; j++)
			{
				if (i != j)
				{
					//倒水的下一步
					tmp = cur; tmp.step++;
					pour(i, j);
					
					if (!vis[tmp.v[1]][tmp.v[2]][tmp.v[3]])
					{
						q.push(tmp);
						vis[tmp.v[1]][tmp.v[2]][tmp.v[3]] = 1;
					}
				}
			}
		}
	}
	printf("NO\n");
	return;
}

int main()
{
	while (scanf("%d%d%d", &S, &M, &N) != EOF && (S || M || N))
	{
		memset(vis, 0, sizeof(vis));
		if (M < N)	swap(M, N);
		V[1] = S; V[2] = M; V[3] = N;
		bfs();
	}
	return 0;
}